# Greetings from The On-Line Encyclopedia of Integer Sequences! http://oeis.org/ Search: id:a237591 Showing 1-1 of 1 %I A237591 #174 Jun 24 2023 12:55:23 %S A237591 1,2,2,1,3,1,3,2,4,1,1,4,2,1,5,2,1,5,2,2,6,2,1,1,6,3,1,1,7,2,2,1,7,3, %T A237591 2,1,8,3,1,2,8,3,2,1,1,9,3,2,1,1,9,4,2,1,1,10,3,2,2,1,10,4,2,2,1,11,4, %U A237591 2,1,2,11,4,3,1,1,1,12,4,2,2,1,1,12,5,2,2,1,1,13,4,3,2,1,1,13,5,3,1,2,1,14,5,2,2,2,1 %N A237591 Irregular triangle read by rows: T(n,k) is the difference between the total number of partitions of all positive integers <= n into exactly k consecutive parts, and the total number of partitions of all positive integers <= n into exactly k+1 consecutive parts (n>=1, 1<=k<=A003056(n)). %C A237591 The original name was: Triangle read by rows: T(n,k) = A235791(n,k) - A235791(n,k+1), assuming that the virtual right border of triangle A235791 is A000004. %C A237591 T(n,k) is also the length of the k-th segment in a zig-zag path on the first quadrant of the square grid, connecting the point (n, 0) with the point (m, m), starting with a segment in vertical direction, where m <= n. %C A237591 Conjecture: the area of the polygon defined by the x-axis, this zig-zag path and the diagonal [(0, 0), (m, m)], is equal to A024916(n)/2, one half of the sum of all divisors of all positive integers <= n. Therefore the reflected polygon, which is adjacent to the y-axis, with the zig-zag path connecting the point (0, n) with the point (m, m), has the same property. And so on for each octant in the four quadrants. %C A237591 For the representation of A024916 and A000203 we use two octants, for example: the first octant and the second octant, or the 6th octant and the 7th octant, etc., see A237593. %C A237591 At least up to n = 128, two zig-zag paths never cross (checked by hand). %C A237591 The finite sequence formed by the n-th row of triangle together with its mirror row gives the n-th row of triangle A237593. %C A237591 The connection between A196020 and A237271 is as follows: A196020 --> A236104 --> A235791 --> this sequence --> A237593 --> A239660 --> A237270 --> A237271. %C A237591 Comments from _Franklin T. Adams-Watters_ on sequences related to the "symmetric representation of sigma" in A235791 and related sequences, Mar 31 2014. (Start) %C A237591 The place to start is with A235791, which is very simple. Then go to A237591, also very simple, and A237593, still very simple. %C A237591 You then need to interpret the rows of A237593 as Dyck paths. This interpretation is in terms of run lengths, so 2,1,1,2 means up twice, down once, up once, and down twice. Because the rows of A237593 are symmetric and of even length, this path will always be symmetric. %C A237591 Now the surprising fact is that the areas enclosed by the Dyck path for n (laid on its side) always includes the area enclosed for n-1; and the number of squares added is sigma(n). %C A237591 Finally, look at the connected areas enclosed by n but not by n-1; the size of these areas is the symmetric representation of sigma. (End) %C A237591 From _Hartmut F. W. Hoft_, Apr 07 2014: (Start) %C A237591 The row sum is A235791(n,1) - A235791(n,floor((sqrt(8n+1)-1)/2)+1) = n - 0. %C A237591 Mathematica function has been written to check the conjecture as well as non-crossing zig-zag paths (Dyck paths rotated by 90 degrees) up through n=30000 (same applies to A237593). (End) %C A237591 The n-th zig-zag path ending at the point (m, m), where m = A240542(n). - _Omar E. Pol_, Apr 16 2014 %C A237591 From _Omar E. Pol_, Aug 23 2015: (Start) %C A237591 n is an odd prime if and only if T(n,2) = 1 + T(n-1,2) and T(n,k) = T(n-1,k) for the rest of the values of k. %C A237591 The elements of the n-th row of triangle together with the elements of the n-th row of triangle A261350 give the n-th row of triangle A237593. %C A237591 T(n,k) is also the area (or the number of cells) of the k-th vertical side at the n-th level (starting from the top) in the left hand part of the front view of the stepped pyramid described in A245092, see Example section. %C A237591 (End) %C A237591 From _Omar E. Pol_, Nov 19 2015: (Start) %C A237591 T(n,k) is also the number of cells between the k-th and the (k+1)st line segments (from left to right) in the n-th row of the diagram as shown in Example section. %C A237591 Note that the number of horizontal line segments in the n-th row of the diagram equals A001227(n), the number of odd divisors of n. (End) %C A237591 Conjecture: the values f(n,k) in the n-th row of the triangle are either 1 or 2 for all k with ceiling((sqrt(4*n+1)-1)/2) <= k <= floor((sqrt(8*n+1)-1)/2) = r(n), the length of the n-th row, though the lower bound need not be minimal; tested through 2500000. See also A285356. - _Hartmut F. W. Hoft_, Apr 17 2017 %C A237591 Conjecture: T(n,k) is the difference between the total number of partitions of all positive integers <= n into exactly k consecutive parts, and the total number of partitions of all positive integers <= n into exactly k+1 consecutive parts. - _Omar E. Pol_, Apr 30 2017 %C A237591 From _Omar E. Pol_, Aug 31 2021: (Start) %C A237591 It appears that T(n,2)/T(n,1) converges to 1/3. %C A237591 It appears that T(n,3)/T(n,2) converges to 1/2. %C A237591 It appears that T(n,4)/T(n,3) converges to 3/5. %C A237591 It appears that T(n,5)/T(n,4) converges to 2/3. (End) %C A237591 In other words: T(n,k) is the length of the k-th line segment of the largest Dyck path of the symmetric representation of sigma(n). - _Omar E. Pol_, Sep 08 2021 %H A237591 G. C. Greubel, Table of n, a(n) for the first 150 rows %F A237591 T(n,k) = ceiling((n+1)/k - (k+1)/2) - ceiling((n+1)/(k+1) - (k+2)/2), for 1 <= n and 1 <= k <= floor((sqrt(8n+1)-1)/2). - _Hartmut F. W. Hoft_, Apr 07 2014 %e A237591 Triangle begins: %e A237591 1; %e A237591 2; %e A237591 2, 1; %e A237591 3, 1; %e A237591 3, 2; %e A237591 4, 1, 1; %e A237591 4, 2, 1; %e A237591 5, 2, 1; %e A237591 5, 2, 2; %e A237591 6, 2, 1, 1; %e A237591 6, 3, 1, 1; %e A237591 7, 2, 2, 1; %e A237591 7, 3, 2, 1; %e A237591 8, 3, 1, 2; %e A237591 8, 3, 2, 1, 1; %e A237591 9, 3, 2, 1, 1; %e A237591 9, 4, 2, 1, 1; %e A237591 10, 3, 2, 2, 1; %e A237591 10, 4, 2, 2, 1; %e A237591 11, 4, 2, 1, 2; %e A237591 11, 4, 3, 1, 1, 1; %e A237591 12, 4, 2, 2, 1, 1; %e A237591 12, 5, 2, 2, 1, 1; %e A237591 13, 4, 3, 2, 1, 1; %e A237591 13, 5, 3, 1, 2, 1; %e A237591 14, 5, 2, 2, 2, 1; %e A237591 14, 5, 3, 2, 1, 2; %e A237591 15, 5, 3, 2, 1, 1, 1; %e A237591 ... %e A237591 For n = 10 the 10th row of triangle A235791 is [10, 4, 2, 1] so row 10 is [6, 2, 1, 1]. %e A237591 From _Omar E. Pol_, Aug 23 2015: (Start) %e A237591 Illustration of initial terms: %e A237591 Row _ %e A237591 1 _|1| %e A237591 2 _|2 _| %e A237591 3 _|2 |1| %e A237591 4 _|3 _|1| %e A237591 5 _|3 |2 _| %e A237591 6 _|4 _|1|1| %e A237591 7 _|4 |2 |1| %e A237591 8 _|5 _|2 _|1| %e A237591 9 _|5 |2 |2 _| %e A237591 10 _|6 _|2 |1|1| %e A237591 11 _|6 |3 _|1|1| %e A237591 12 _|7 _|2 |2 |1| %e A237591 13 _|7 |3 |2 _|1| %e A237591 14 _|8 _|3 _|1|2 _| %e A237591 15 _|8 |3 |2 |1|1| %e A237591 16 _|9 _|3 |2 |1|1| %e A237591 17 _|9 |4 _|2 _|1|1| %e A237591 18 _|10 _|3 |2 |2 |1| %e A237591 19 _|10 |4 |2 |2 _|1| %e A237591 20 _|11 _|4 _|2 |1|2 _| %e A237591 21 _|11 |4 |3 _|1|1|1| %e A237591 22 _|12 _|4 |2 |2 |1|1| %e A237591 23 _|12 |5 _|2 |2 |1|1| %e A237591 24 _|13 _|4 |3 |2 _|1|1| %e A237591 25 _|13 |5 |3 _|1|2 |1| %e A237591 26 _|14 _|5 _|2 |2 |2 _|1| %e A237591 27 _|14 |5 |3 |2 |1|2 _| %e A237591 28 |15 |5 |3 |2 |1|1|1| %e A237591 ... %e A237591 Also the diagram represents the left part of the front view of the pyramid described in A245092. For the other half front view see A261350. For more information about the pyramid and the symmetric representation of sigma see A237593. (End) %e A237591 From _Omar E. Pol_, Sep 08 2021: (Start) %e A237591 For n = 12 the symmetric representation of sigma(12) in the fourth quadrant is as shown below: _ %e A237591 | | %e A237591 | | %e A237591 | | %e A237591 | | %e A237591 | | %e A237591 _ _ _| | %e A237591 _| _ _| %e A237591 _| | %e A237591 | _| %e A237591 | _ _|1 %e A237591 _ _ _ _ _ _| | 2 %e A237591 |_ _ _ _ _ _ _|2 %e A237591 7 %e A237591 . %e A237591 The lengths of the successive line segments from the first vertex to the central vertex of the largest Dyck path are [7, 2, 2, 1] respectively, the same as the 12th row of triangle. (End) %t A237591 row[n_]:= Floor[(Sqrt[8*n+1] -1)/2]; f[n_,k_]:= Ceiling[(n+1)/k-(k+1)/2] - Ceiling[(n+1)/(k+1)-(k+2)/2]; %t A237591 Table[f[n,k],{n,1,50},{k,1,row[n]}]//Flatten %t A237591 (* _Hartmut F. W. Hoft_, Apr 08 2014 *) %o A237591 (PARI) row235791(n) = vector((sqrtint(8*n+1)-1)\2, i, 1+(n-(i*(i+1)/2))\i); %o A237591 row(n) = {my(orow = concat(row235791(n), 0)); vector(#orow -1, i, orow[i] - orow[i+1]);} \\ _Michel Marcus_, Mar 27 2014 %o A237591 (Python) %o A237591 from sympy import sqrt %o A237591 import math %o A237591 def T(n, k): return int(math.ceil((n + 1)/k - (k + 1)/2)) - int(math.ceil((n + 1)/(k + 1) - (k + 2)/2)) %o A237591 for n in range(1, 29): print([T(n, k) for k in range(1, int((sqrt(8*n + 1) - 1)/2) + 1)]) # _Indranil Ghosh_, Apr 30 2017 %Y A237591 Row n has length A003056(n) hence column k starts in row A000217(k). %Y A237591 Row sums give A000027. %Y A237591 Column 1 is A008619, n >= 1. %Y A237591 Right border gives A042974. %Y A237591 Cf. A000203, A001227, A024916, A196020, A235791, A236104, A237048, A237270, A237271, A237593, A239660, A239931-A239934, A240542, A244580, A245092, A249351, A259176, A259177, A261350, A261699, A262626, A285356, A286000, A286001. %K A237591 nonn,tabf %O A237591 1,2 %A A237591 _Omar E. Pol_, Feb 22 2014 %E A237591 3 more rows added by _Omar E. Pol_, Aug 23 2015 %E A237591 New name from a comment dated Apr 30 2017. - _Omar E. Pol_, Jun 18 2023 # Content is available under The OEIS End-User License Agreement: http://oeis.org/LICENSE